Momentum space representation of the hydrogen atom Schroedinger equation

The Schroedinger equation for the hydrogen atom is

$$\Big(-\frac{\hbar^2}{2m} \Delta - \frac{e^2}{r} \large)\Psi(\mathbf{r}) = E \; \Psi(\mathbf{r}).$$

I have found that the momentum representation of the above equation reads

$$\frac{\mathbf{p}^2}{2m}\Psi(\mathbf{p}) -\frac{e^2}{2\pi^2 h} \int \frac{\Psi(\mathbf{p'})\mathrm{d}^3\mathbf{p'}}{|\mathbf{p}-\mathbf{p'}|^2} = E \; \Psi(\mathbf{p}).$$ How to derive it? The Fourier transforms of the Laplace operator part and the right hand side are clear. But how about the potential part?

The problem here is the mistaken impression that the momentum representation is connected to the position representation by a Fourier Transform. Contrary to what is said in textbooks, this is only true for Cartesian Coordinates. For curvilinear coordinates (e.g. r, theta, phi) a different transformation is needed. Check out the article: The Hydrogen Atom in the Momentum Representation, John R. Lombardi, Phys. Rev. A, 22, 797 (1980), in which the correct transform leads to a momentum space where all the momentum variables are chosen to be properly conjugate to the three position space coordinates.

• We do have MathJax here to format equations. You can see the notation page in help center for details, if you aren't familiar with it (though it's similar to LaTeX). – Kyle Kanos Jun 6 at 20:29
• Hi, thank you for the wonderful reference I had not known about. – DanielC Jun 6 at 21:22
• I will look at the suggested reference, but I think that these argumemts are not really sound as the equations in OP are not linked with a particular (curvilinear) coordinates system. IMO @Blazej's answer is totality correct. – Jhor Jun 11 at 7:46

You take the Fourier transform of Schrödinger equation in position space. Transformation of the term with potential can be calculated using convolution theorem (Fourier transform of a product is convolution of Fourier transforms - details are easy to find online and the proof is rather straightforward). Fourier transform of $V$ itself can be easily calculated using three facts (I write up to constant factors which can be easily checked):

1. Fourier transform of Laplace operator acting on function is Fourier transform of that function times $-k^2$ (here $k$ is wavector).

2. Laplace operator on $V$ is delta.

3. Fourier transform of delta is a constant.

• Why do I need Laplace operator on V if Laplace is acting only on the wave function? By using the convolution theorem it seems that what I need is the Fourier transform of 1/r. Then I would use the fact that the Fourier transform of a product is the convolution of the respective Fourier transforms. That should solve the problem. – wondering Jan 19 '17 at 12:11
• Yes. I was hinting to the fact that knowing that Laplace operator acting on $1/r$ gives delta you can immediately deduce what is its Fourier transform. – Blazej Jan 19 '17 at 13:46